Question

$$\int \frac{x}{\sqrt{x^2+6x+10}}\text{ dx }$$

Answer 1

 

$$\text{ Let I }=\int \frac{xdx}{\sqrt{x^2+6x+10}}$$
$$x=\text{ A }\frac{d}{dx}(x^2+6x+10)+B$$
$$x=A(2x+6)+B$$
$$x=\left(2A\right)x+6A+B$$
$$\text{Equating Coefficients of like terms}$$
$$2A=1$$
$$A=\frac{1}{2}$$
$$6A+B=0$$
$$6\times \frac{1}{2}+B=0$$
$$B=-3$$
${\displaystyle I = \int \frac{x dx}{\sqrt{x^2+6x+10}}}$
$$=\int \left(\frac{\frac{1}{2}(2x+6)-3}{\sqrt{x^2+6x+10}}\right)dx$$
$$=\frac{1}{2}\int \frac{(2x+6)dx}{\sqrt{x^2+6x+10}}-3\int \frac{dx}{\sqrt{x^2+6x+3^2-3^2+10}}$$
$$=\frac{1}{2}\int \frac{(2x+6)dx}{\sqrt{x^2+6x+10}}-3\int \frac{dx}{\sqrt{{(x+3)}^2+1^2}}$$
$${\text{ let x }}^2+6x+10=t$$
$$\Rightarrow (2x+6)dx=dt$$
$$I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}-3\int \frac{dx}{\sqrt{{(x+3)}^2+1}}$$
$$=\frac{1}{2}\times 2\sqrt{t}-3\text{ log }|x+3+\sqrt{{(x+3)}^2+1}|+C$$
$$=\sqrt{t}-3\text{ log }|x+3+\sqrt{x^2+6x+10}|+C$$
$$=\sqrt{x^2+6x+10}-3\text{ log }|x+3+\sqrt{x^2+6x+10}|+C$$